In $\mathrm{A}_{\mathrm{x}} \mathrm{B}_{\mathrm{y}}$ crystal structure, $\mathrm{A}$ atoms occupied all…
In $\mathrm{A}_{\mathrm{x}} \mathrm{B}_{\mathrm{y}}$ crystal structure, $\mathrm{A}$ atoms occupied all octahedral as well as all tetrahedral voids and B atoms are at FCC centres. What is the formula of compound $\mathrm{A}_{\mathrm{x}} \mathrm{B}_{\mathrm{y}}$.
$\mathrm{AB}_3$
$\mathrm{A}_{10} \mathrm{~B}_3$
$\mathrm{A}_{15} \mathrm{~B}_{36}$
$\mathrm{A}_3 \mathrm{~B}$
Solution
Number of octahedral voids $=\mathrm{x}$ (where $\mathrm{x}=$ total number of atoms in the rinit cell)
Number of tetrahedral Voids $=2 \mathrm{x}$
$\Rightarrow$ Total number of ' $A$ ' atoms $=x+2 x=3 x$
Since all 'B' atoms occupy all the FCC centres:-
$\Rightarrow$ Total number of 'B' atoms $=\left[6 \times \frac{1}{2}\right]+\left[8 \times \frac{1}{8}\right]=4$
$\Rightarrow \quad x=4$ so number of ' $A$ ' atoms $=3 x=3 \times 4=12$
So, the formula will be $A_{12} B_4$ or $A_3 B$